An isosceles triangle is inscribed in a circle, with . Show that the altitude from to passes through the centre of the circle.

Question

Homi · Accepted Answer

Understand the Question

An isosceles triangle $\triangle ABC$ with $AB = AC$ is inscribed in a circle, and $AD$ is the altitude drawn from $A$ to $BC$ .
A key property of circles is that the perpendicular bisector of any chord passes through the centre of the circle.
To prove that the altitude $AD$ passes through the centre, we first show that $AD$ bisects the chord $BC$ using triangle congruence, establishing $AD$ as the perpendicular bisector of $BC$ .

Step 1 · Show that Altitude $AD$ Bisects $BC$

Let $AD \perp BC$ be the altitude drawn from vertex $A$ to base $BC$ . Diagram 1

In right triangles $\Delta ABD$ and $\Delta ACD$ :

$\angle ADB = \angle ADC = 90^\circ \quad (\text{Since } AD \perp BC)$
$AB = AC \quad (\text{Given})$
$AD = AD \quad (\text{Common side})$

By the RHS congruence rule: $\Delta ABD \cong \Delta ACD$

Therefore, by CPCT: $BD = DC$

Thus, $AD$ bisects the side $BC$ .

Step 2 · Prove $AD$ Passes Through the Centre

Since $AD \perp BC$ and $BD = DC$ , $AD$ is the perpendicular bisector of $BC$ .

Since $BC$ is a chord of the circle and the perpendicular bisector of any chord passes through the centre of the circle, $AD$ must pass through the centre.

Hence, the altitude from $A$ to $BC$ passes through the centre of the circle.

Answer

Hence proved that the altitude from $A$ to $BC$ passes through the centre of the circle.

Common Mistakes

Assuming Bisector Directly: Do not assume $AD$ bisects $BC$ without proof; you must establish $\Delta ABD \cong \Delta ACD$ using the RHS congruence criterion first.
Chord Property Application: Remember that the perpendicular bisector of any chord of a circle always passes through its centre.

Question 2

More questions in Exercise 5.3